AP = 64, 60, 56, …
Here, a = 64, d = 60 – 64 = -4
⇒ 1088 = n[132 – 4n]
⇒ 4n2 – 132n + 1088 = 0
⇒ n2 – 33n + 272= 0
⇒ n2 – 16n - 17n + 272 = 0
⇒ n(n – 16) - 17(n – 16) = 0
⇒ (n – 16)(n – 17) = 0
⇒ n – 16 = 0 or n – 17 = 0
⇒ n = 16 or n = 17
If n = 16, a = 64 and d = -4
a16 = 64 + (16 – 1)(-4)
a16 = 64 + 15 × -4
a16 = 64 – 60
a16 = 4
and If n = 17, a = 64 and d = -4
a17 = 64 + (17 – 1)(-4)
a17 = 64 + 16 × -4
a17 = 64 – 64
a17 = 0
Now, we will check at which term the sum of the AP is 544.
⇒ S17 = 17 × 32
⇒ S17 = 544
So, the terms may be either 17 or 16 both holds true.
We get a double answer because the 17th term is zero and when we add this in the sum, the sum remains the same.